wanminliu@gmail.com
20210920
.
If
✗
=
0
,
then
②
holds
.
So
solutions
are
given
by
10
,
Y
)
,
YHR
That
is
,
the
stationary
points
are
given
by
solution
.
the
set
{
lay
,
/
year
}
(A)
Leta
__
0
,
we
hav e
fix
.y
,
=
-
✗
'
y
}
-
✗
2
-
f-
✗
IX.
7)
=
-
zxy
}
-
2
✗
=
-2×(93+1)
✗
fy
IX.
y
)
=
-
✗
2
.
3yd
We
want
to
solve
ix.
y
,
for
the
equation
$&
:
③
Tf
(
✗
-
Y
)
=
8
zero
vector
in
,pz
f×(
✗
'
41=-2×(43+1)
fycxiy
]
=
Day
-
3×42
=
7189-3×7
)
④
ie
.
(-2×47-1)
,
-
X'
34
'
)
=
10,0
)
we
want
solve
pfcxip
-_
o
3×32=0
①
that
is
.
I
{
-2×(5+1)=0
⑤
2×(73+1)
=
0
②
y
Ifa
-3×47=0
⑥
①
:
✗
=
0
or
y
=
0
If
7=0
,
then
②
z×=o
11=0
.
wanminliu@gmail.com
⑤
y
3=-1
7=-1
⑤
y
3=-1
4=-1
or
✗
=
0
or
✗
=
0
•
if
✗
=
0
,
then
⑥
Y
-
8a=0]µ=o
.
•
If
✗
=0
,
then
⑥
Y
-
8
a
=
0
since
a
>
o
(
Ato
)
⑤
y=o
•
4-
1×+03
.
7=-1
,
put
it
into
⑥
So
a
solution
is
(
Op
)
(4)
(
8A
-3×4-11=0
•
If
✗
too
y=
-1
,
then
8h
-13×2=0
since
a⇐
⑥
C-
1)
(
fat
3×-1 =0
ther e
is
no
solution
for
✗
.
✗
2=8
,
c-
a)
>oa
so
the
only
stationary
pint
is
(
0,07
.
✗
=±§Ñ
So
stationery
points
are
10.01
,
LIFE
!
-
1)
,
1-
P
,
-
1)
.
wanminliu@gmail.com
(d)
a=
-1
:
fix
,y)=
-45-573
-
X
'
f-
✗
(
x
.
-111
ftp.9-0
near
10 ,0
)
=
Po
upto
order
2
10 ,01
=
°
In
general
,
let
8=4,7
)
,
then
upto
order
?
fylx.in/@..,= opzCX.YY--flx.-Dl-fxlxiD/
(
X
-
X
.
)
Po
Po
Po
⑦
f××
=
-243-1
)
1-
fy
IX.
7)
/
poly
-
Yo
)
So
fxxlx .vn/,qg---z......1+..a...,.......
*
↳
=
.ws
,
Po
fxylx.is/ @.g--0-2fxylX.Y)/polX-Xao)G-Yo
)
⑧
fyy
=
-
(8+3×4)
-
y
(3×2)
1-
fyylx.us /p.oly-Yoi)fyylXiD/go.g=
-8
Now
{
f-
✗
1×-41=-2×(93+1)
⑦
fyexiy
,
=
-418+3×4
)
⑧
↳
Paki
=
0+0×+07-1
#
(-2×40×7+6,5)
=
-1×4×5
)
☐
wanminliu@gmail.com
so
the
domain
D
is
the
region
inside
or
on
the
spher e
of
radius
V2
,
Mth
center
✗
z
10,903
,
and
below
or
on
the
plane
2-
=
+1
.
r
Sol
.
(
method
from
Geometry
)
(
1.1
.
1)
=P
The
inequality
✗
'
+
y
'
-12-2
£2
fg"#
>
y
gives
a
ball
centered
at
origin
with
rir dius
V2
and
its
inside
.
±
The
Inequality
Z
E
t
sires
the
2-
=/
plane
and
the
space
"
below
"z=i
.
The
fourth
f-
ix.
y
.z
,
is
the
Square
of
distant
✗
2-
from
a
point
in
D
to
the
r
the
point
Po
=
(
1.
1
.
17
.
r
.
>
Y
Taking
the
line
passing
through
pond
0
r
vhih
intersect
the
sphere
at
two
points
✗
[
A
&
B
wanminliu@gmail.com
points
online
hav e
coontmot
center
of
the
sphere
,
d-
the
¥¥÷É÷
"
"
"
"
"
:*
.
IX.
X
-
X
)
intersection
of
sphere
and
plane
is
a
big
circle
,
with
rad in
or
throug h
OP
☐
and
internet
dist
(
o
,
D)
=
1412€
=D
ix.
✗
-
✗
1
the
sphere
.
debt
10
B)
=
rz
go
dirt
(
BD
=
rz
-103
On
the
line
op
,
point
has
coordinate
LX
,
X
.
X
)
.
So
A
&
B-
satisfy
distlB.DE
@
+
E)
"
✗
4×4×1=2
✗
=±§
A-
=
(
Fi
,
F)
and
=5
-128
B-
-
1-
E.
if
,
-
E)
=
fmax
wanminliu@gmail.com
solution3.2-flag.multipk.ir
method
)
Remark
:
°""""""
D=
{
IX.
7.
z
,
/
✗
¥242
;
2-
El
}
•
Max
f-
IX.
y
,
Z
)
=
Max
f-
ix.
7.
2-
I
f-
ix.
y
.
E)
=
☒
-1144-132+(2--1)
'
spher e
✗
2+5+5=2
w.eu#-tw-Mmax-ttI
"
then
we
can
use
a
simpler
by
the
hag
.
multiplier
method
.
hey
.
multiplier
,
that
:)
Define
glx.y.zj-XY-yt-E-LL lx.y.tn
)=
f-
+
✗
g
h
IX.
Y
.
7)
=
2-
-
I
use
a
little
geometry
to
argue
this
.
Then
LIX.Y.tt
.µ)=
f-
+
✗
g
+
µ
h
L×=f×
-1×4=21×-4
"
"
stepl
.
cheek
critical
pts
Ly
=
fy
+
✗
gy
=
24-1
)
+
✗
ZY
Lz
=
fz
+
xgz
-1
µ
=
zcz
-
1)
1-
HE
1-
µ
↳
=
g
=
✗
2+-14-5-2
TL
=
§
gives
us
her
=
h
=
2-
-
I
wanminliu@gmail.com
[
✗
=
f-
✗
1-
✗
9×
=
21×-11
+
✗
2X
=
0
①
Ly
=
fyxxgy
=
24-1
)
+
✗
ZY
=
0
②
{
↳
=p
,
+
✗
g
,
+
µ
=
zezy
,
+
✗
*
+
µ
=
☐
③
⑧
↳
=
'
=
"
"
"
"
"
"
=
°
④
1)
her
=
h
=
z
-
I
=
o
⑤
a
BB
⑤
z
✗
45=1
④
§
,
✗
=y=±§
①
&
②
✗
=y
→
critical
points
II.
E.
aorta
is
the
critical
parts
ar e
on
the
bounty
step3_Chehbunihypts-
of
the
region
)
evaluate
the
fourth
f.
↳
flx.Y.tt
)
/
(
§
,
#
,
,
,
=/
I
-154€
-42
M¥
dbk
2=1
✗
45--2-1=1
f-
Kitty
y
=
+
1)
'
+
+15
or
wanminliu@gmail.com
{
✗
4-71-2-2=2
maxflx.ie#
2-
£
I
•
we
now
compute
on
the
boundary
A-
ix.
yj-ttfcx.ie
.
z-x2#
)
we
do
the
cae
xtyt.az
=
☒
-47-4-07
'
so
✗
45+2=-2
¥
Now
Ofz
'
£1
YE
-
Z
's
0
PEZ
-
£2
E
2
a.
,
*
www.am.a.m.ae
CAsl
Off
£1
region
{
ix.
y
,
HRT
of
✗
4-
Y'
£2
}=
region
M
•
we
compute
the
boundary
Itis
no
harm
Bay
)
f-
ix.
y
,
-
TEY
)
totakegefualithowe
,
CM
-152<-2
f
o
=
X-D
4-
g-
1)
7-
(2×1%+1)
'
NOW
of
2-
2£
2
f-
-2
f-
72
go
2-
=
-
¥-5
Of
2-2-2
<
2
we
compute
the
minx
Blx
-
y
)
in
the
wanminliu@gmail.com
region
of
me
find
that
{
IX.
4)
FRY
of
X'
+
y
'
£2
}
=
region
N
the
third
term
of
Btx
.
Y
)
(2-1×475-+1)
2
is
always
bigger
than
the
third
term
of
A
ix.
y
)
,
that
is
i
.*
.
So
we
find
that
Max
B
Hill
3
Max
A-
ix.
Y
)
region
N
region
µ
To
compute
the
mail
of
fix.y.to
)
we
only
need
to
compute
it
on
the
region
N
for
BH
.
Y
)
.
wanminliu@gmail.com
we
only
compute
Blxiy
then
N
Bay
)
f-
ix.
y
,
-
FEY
)
=
X-D
4-
g-
1)
7-
(2×1%+1)
'
B.
✗
=
21×-17-12
(2-+2-42/+1)
.
-21.12
-
✗
'
-55
?
C-
zx
,
By
=
24-1
)
-12C
2-x
+
1)
I
(
2-
✗
Zyy
-
I
1-
zyj
Find
the
ordeal
put
for
B.ie/Bx=o0By--0
②
①
✗
Y
-
2.
☒
✗
21×-117=2
(
Y
-
1)
✗
✗
=Y
.
③
put
③
Into
①
21×-13-2
(2-2%+1)
(2-2×2)
-
I
✗
=
o
ex
-
1)
=
✗
(
I
+
(2-2×2)
-
£
)
=
-1
{
✗
•
✗
'
=
2-2×2
✗
=
So
ix.
7)
=
1-
§
,
-
)
.
So
2-
=
-
z-x'Ñ=
-
§
Max
Bail
)
=
Bl
-
§
,
-
t-2-f-IY-fz-I.IT
+15--31-5+15=5+256
.
N
wanminliu@gmail.com
We
still
need
to
compute
'
¥0
disk
2=1
the
call
3.
I
XEi=
2-1 =1
U
Ccxiyjtt
1×-114
y
-
1)
7-
o
'
on
the
Domain
k-ttllx.us/oex4-y
's
,
}
It
is
easy
to
know
Max
Clay
)
£
Max
(
☒
-15+4-1
)I(2-×
D)
=/
max
Bcxi
)
region
K
region
K
¥É
region
N
So
we
don't
need
to
compute
Max
C
ix.
y
)
.
region
K
wanminliu@gmail.com
In
summary
,
for
critical
points
.
Since
II.
Fz
,
1)
or
f-
if
1)
5+256
>
+42>2
(
tf
-42
We
have
f-
"
'
"
4¥
,
#
,
,
,
=/
§y4y§yz
The
ma
"
value
of
f
is
obtained
at
the
point
-45
,
-
§
,
-
E)
with
Max
-1
Value
5+256
.
f-
"
"
"
I
y
=
+
1)
'
+
+15
-☐
Remand
.
Usin g
the
same
method
,
we
For
the
boundary
point
fjf
,
-
§
,
-
§
)
can
argue
that
(
from
Case
3.
2.
2)
min
f-
IX.
Y
.
2)
=
min
Akil
)
we
have
D
M
f-
IX.
Y.tl/fq
,
-
§
,
.gg
,
=
5+28
.
=
⑤
-
ri
☐
